CHEMISTRY COMMUNITY

Ashley Zhu 1A

Concentration is calculated by the equation moles/liters, so as long as you find out how many moles there are of each gas and divide by the volume of the container, in this case a 0.250 L flask, then you will be able to find the concentration of the gases. You are given the mass of CO2 gas at equili...

Ashley Zhu 1A

The two equations you have there are essentially the same equations derived from delta(U) = delta(H) + w where w is work. Since w = -Pdelta(V) where P is pressure and delta V is the change in volume, we can rewrite the equation as delta(U) = delta(H) - Pdelta(V). By the ideal gas law, PV = nRT, -Pde...

Ashley Zhu 1A

This problem states that it is a first order decomposition of SO2Cl2, so you know that the only reactant is SO2Cl2 because that's what decomposition means. All parts of the question don't actually require the use of a complete chemical equation for the decomposition, so you don't actually need to fi...

Ashley Zhu 1A

Not sure yet, but Dr. Lavelle said he'd be talking about it in lecture on Friday

Ashley Zhu 1A

The problem tells you that N2O5 decomposes by a first order reaction, so the rate law will have the form rate = k[A] where A is the reactant, N2O5, and the it is raised to the first power since it is a first order reaction. You are also given k = 5.2 x 10^-3. As mentioned before, you don't need to d...

Ashley Zhu 1A

Dr. Lavelle said in class that we won't have to worry about third order reactions, but for first and second order reactions, you can plot the data given to you on a graph or examine the rate law. If a graph of ln[A] vs. t shows a straight line with a negative slope, you know that it is a first order...

Ashley Zhu 1A

the half reaction for the anode here should be H2O --> O2 + H+ and you first balance the O's and then H's and then the charge to get 2H2O --> O2 + 4H+ + 4e- ; this way you see that the O has been oxidized

Ashley Zhu 1A

you do not need to add Pt(s) to liquids in cases where you have Hg(l) and Hg ions, but I am not sure if there are any exceptions/other cases

Ashley Zhu 1A

I think the solution manual is correct in saying that O2 and H+ are the reactants in the oxidation half reaction because there is no way to get H+ ions from O2 gas, so they must react together to form water

Ashley Zhu 1A

by separating the equation, Dr. Lavelle means that you should separate the oxidation part from the reduction part for example, if you have the equation Cu2+ + Zn --> Cu + Zn2+ you can separate the Cu's and Zn's to get Cu2+ + 2e- --> Cu (reduction) and Zn --> Zn2+ + 2e- (oxidation) (you don't see the...

Ashley Zhu 1A

the cathode will contain the reduction reaction and the anode will contain the oxidation reaction. something I find helpful to help me remember is "red cat" for reduction in cathode and "an ox" for anode = oxidation

Ashley Zhu 1A

usually, a more massive atom will have greater entropy because that means it has more protons, neutrons, and electrons than anther atom with less mass, and because it has more particles, there are more possible states it could be in, increasing its total entropy

Ashley Zhu 1A

Cp,m is for an ideal monoatomic gas at constant pressure and Cv,m is for an ideal monoatomic gas at constant volume. You use these values usually when you are trying to calculate the change in entropy when there is a temperature change.

Ashley Zhu 1A

For a reversible reaction, the system is at equilibrium and when it is either compressed or expanded, it does so slowly in infinitely small increments, and you use the equation w = -nRTln(V2/V1) for work. For an irreversible reaction, the reaction is happening under a constant external pressure Pext...

Ashley Zhu 1A

If a system is doing work, it is usually expanding unless the problem specifically states that the system is at constant volume.

Ashley Zhu 1A

for part a), when you plugged in the numbers into w = -PdV you get -2.23 atm x L (single digits) but work is in terms of joules, so you need to convert that into joules using 1 atmL = 101.325 J, which gives you -226 J for your answer the same thing needs to be done for part b) unless for R you use 8...

Ashley Zhu 1A

when calculating the heat of the ice cube, why do we use the specific heat of water in its liquid form and not its solid form?

Ashley Zhu 1A

I'm assuming you're referring to this equation w = - ∫ PdV (limits of integration from V1 to V2) = -nRT ln (V2/V1), in which case the ln comes from taking the integral of 1/V when you replace P with nRT/V (ideal gas equation rearranged) and the constants are taken out so you're left with ∫ dV/V. Sin...

Ashley Zhu 1A

Gaining an H atom does not mean gaining an H+ ion since an H atom still has an e-; this is an important distinction to make because when that new H atom is gained, its one e- will spend more time around the central atom it bonded to because it is more electronegative, and thus, that central atom is ...

Ashley Zhu 1A

since the standard enthalpy of formation is the delta H for the formation of one mole of a substance from its elements in their most stable form, by definition, the standard enthalpy of formation of elements in their most stable form is = 0. the most stable form of oxygen is O2, so since the delta H...

Ashley Zhu 1A

open systems are usually exposed to the air and can exchange both matter and energy: in this case, the gas burning in a car engine (releases heat, which is part of energy and exhaust which contains matter) and a living plant (exchanges gases like O2 and CO2 which counts as matter) are open systems c...

Ashley Zhu 1A

Since Kc = 5.66 x 10^-10 <<< 10^-3, you can disregard the change in molarity caused by x for SO2 and O2, so instead of having Kc = (2x)^2/(0.522-x)^2(0.633-x), you'll just have Kc = (2x)^2/(0.522)^2(0.633), which you won't even have to use the quadratic equation for.

Answer should be D.

Ashley Zhu 1A

it's always best to save rounding for the end, and this could mean either the equilibrium constant value or the x-value, depending on what the question is asking for; basically, only round for whichever number is your final answer.

Ashley Zhu 1A

In this problem, the water in the equation is in its gaseous state, so we need to include it in the ICE table. The only time we don't include water is when it is a liquid/is acting as a solvent.

Ashley Zhu 1A

the same trend also applies for pKb: the lower the pKb, the stronger the base

Ashley Zhu 1A

You could also potentially change K by multiplying the reaction equation by a whole number. If you double the stoichiometric coefficients of everything in the equation, the K will become K^2, which will either increase K is K > 1 or decrease K if K < 1.

Ashley Zhu 1A

You are given that ozone forms oxygen in the equilibrium reaction 2O3 -->/<-- 3O2. It gives you different conditions, one where 0.10 mol of ozone is put in a 1.0L vessel and then another where 0.50 mol is put in a 1.0L vessel, both of which are at the same temperature. This means that the equilibriu...

Ashley Zhu 1A

I think the way the textbook explains it is that K and Kp are essentially the same thing, using partial pressures for reactions that have gaseous compounds while Kc is specific to a K that uses concentrations.
Pages 424 and 437 in the 6th edition might be helpful if you want to look at it.

Ashley Zhu 1A

What's going on in the flask is the reaction X2 --> 2X, and normally, you'd just use the equation K = (Px)^2/(Px2). However, for this, I used the partial pressures of each gas since it only gives you the pressure of X2 at the very beginning, so you modify the equation to be K = (nx*Pi)^2/(nx2*Pi) wh...

Ashley Zhu 1A

since the given chemical equation is all gases, you can rearrange the pv = nrt equation to p = nrt/v = Mrt (where M is molarity for concentration since n/V is mol/L) to find the partial pressures of the gases from the given concentrations; then, you can find the equilibrium constant using P(products...

Ashley Zhu 1A

CH3CHO2- is the base and CH3CO2H is its conjugate acid (first pair). H2O is the acid and OH- is its conjugate base (second pair).

Ashley Zhu 1A

Whether H2O will act as an acid or base depends on the nature of the compound it is reacting with. For example, if water were to react with ammonia (a weak base), then it would be an acid: NH3 + H2O --> NH4+ + OH- (water is donating its proton and ammonia is accepting it). If water were to react wit...

Ashley Zhu 1A

They're essentially the same thing. Cyano is just the older naming convention and cyanido is the newer name agreed on by the IUPAC. I think you can use either one and be correct since the book said the old names are still heavily in use.

Ashley Zhu 1A

You know that molecules like ammonia (NH3) have a zero charge and compounds like cyanide (CN-) have charges on them. Add the charge of the compound/molecule to the oxidation state of the anion (i.e. -1 for Cl). Then, you look at the overall charge of the coordinate compound to figure out the oxidati...

Ashley Zhu 1A

If all those atoms in the same group where bonded to the same atom, let's say the halogen group are all bonded to hydrogen atoms (HF, HCL, HBR, HI), the atom that will form a more polar bond would be the most electronegative one, which is the first atom in each group. This is because it has the leas...

Ashley Zhu 1A

A radical is an atom or a molecule that has an unpaired valence electron such as fluorine or CH3. Radicals are generally very unstable and can cause harm to the body by denaturing macromolecules.

Ashley Zhu 1A

Can a pi bond be formed before a sigma bond is formed?

Ashley Zhu 1A

Why does a p-p orbital overlap form a sigma bond and not a pi bond? What is an example of this?

Ashley Zhu 1A

I would first begin by looking at the nature of the bonds in the molecule and then the molecule shape because the intermolecular forces formed depend on the polarity of the molecule. If the molecule consists entirely of nonpolar bonds, then the molecule itself will be nonpolar as well, meaning it wi...

Ashley Zhu 1A

You're right! The shape wouldn't be linear if there was a/were lone pair(s) on the central atom. That's why for a) where the bond angle is 120 degrees, it is possible for there to be lone pairs on the central atom since the shape of the molecule is bent (repulsion from the lone pair(s) push the two ...

Ashley Zhu 1A

You don't really look at the charge of the anion when determining its polarizability but rather its size and electronegativity. Polarizability increases as the ion gets larger and less electronegative since the electrons aren't held as tightly by the nucleus. One of the homework problems shows this ...

Ashley Zhu 1A

During lecture, Dr. Lavelle said that the partial positive charge on a hydrogen interacts with the/a lone pair on an N, O, or F. I always thought the interaction was with an N, O, or F that had a partial negative charge. Is this incorrect or do both occur at the same time when a hydrogen bond forms?...

Ashley Zhu 1A

here's an example of what nitrate will actually look like in real life. notice that resonance is indicated by the dashed lines, meaning that those bonds are neither single or double bonds but a combination of the two. https://en.wikipedia.org/wiki/Nitrate#/media/File:Nitrate-ion-with-partial-charges...

Ashley Zhu 1A

when you draw the resonance structures, it seems like the bonds are of different lengths since there's a combination of two single bonds and one double bond (nitrate) or one single bond and one double bond. however, in reality, the bonds in the actual polyatomic ions are neither single or double bon...

Ashley Zhu 1A

Also, if you see that your Lewis structure is going to have resonance structures, then you know that all of the bonds involved will have the same length.

Ashley Zhu 1A

If you're interested, you can search up "ozone lewis structure" and you'll see that it has two resonance structures, but in reality it exists as something in between like this: https://upload.wikimedia.org/wikipedia/ ... dipole.png

Ashley Zhu 1A

You can write a Lewis structure with dotted lines between the atoms to show resonance, but I think it'd be safer to write out each resonance structure and have the double arrows in between each to show resonance since that's what Dr. Lavelle did in class.

Ashley Zhu 1A

When the elements that can have an expanded octet form compounds, they are most stable at their lowest energy state. You can determine this by calculating formal charges on each of the elements in the compound. Usually, the Lewis structures with the most 0's/formal charges closest to 0 are the most ...

Ashley Zhu 1A

3 orbitals are present in the 4p subshell since there are always 3 orbitals in any p subshell - the px, py, and pz.

Ashley Zhu 1A

When an e- transitions from the 1s orbital to the 2p orbital, it is moving to a higher energy state in a different shell (from n=1 to n=2). Since the 2s and 2p orbitals are all in the same shell, they all have the same energy in a hydrogen atom due to its degenerate nature.

Ashley Zhu 1A

1 eV = 1.602 x 10^-19 J

you can find this on the constants and equations link on the class website. it'll also be provided on the constants and equation sheet during tests, so no need to worry about memorizing it!

Ashley Zhu 1A

Thank you! Does this rule apply to gold also? And in that case, why is the e- configuration for tungsten(W), which is in period 6 under chromium, [Xe]4f^14 5d^4 6s^2 and not [Xe] 4f^14 5d^5 6s^1? (part e) of the same problem)

Ashley Zhu 1A

A subshell and an orbital refer to the same thing, and a shell includes an orbital/subshell. For example, the second shell (n = 2) would have one s orbital and one p orbital.

Ashley Zhu 1A

Schrodinger's equation uses wave function to describe an e- in an atom. The Schrodinger equation is H(psi) = E(psi) where psi is the pitchfork looking Greek letter representing the wave function, and what the equation is telling us is that basically after taking the double derivative of the wave fun...

Ashley Zhu 1A

Part a) of number 13 asks for the ground-state electron configuration for silver, and the answer says it's [Kr] 4d^10 5s^1. The textbook explicitly stated that only chromium and copper were exceptions to the general electron configuration rules and didn't include silver but the answer suggests other...

Ashley Zhu 1A

The unit "hertz" means 1/seconds, so when you multiply J x s by Hz = 1/s, the seconds cancel out and you're left with J.

Ashley Zhu 1A

MHz means "mega hertz" and is equal to 1,000,000 Hz, so you're right in saying that M is just a prefix. When you do a problem always convert back into the base unit unless the problem specifically tells you otherwise.

Ashley Zhu 1A

There's also a Constants and Equations link on the class website that has the Rydberg constant, Avogadro's number, and a bunch of other constants and useful numbers like the mass of an electron. I'm pretty sure that's the sheet that will be provided to us on tests and exams.

Ashley Zhu 1A

According to the wave model of light, increasing the intensity of the light should increase the energy of the wave since the amplitude of a wave = its energy and intensity is amplitude squared, so the scientists performing the experiment thought that increasing the intensity of the light should allo...

Ashley Zhu 1A

I had a bit of trouble with this problem as well and ended up just using the guess and check method with Equation 2 on page 7. Since you know that it is part of the Lyman Series, n1 has to = 1 and then I just plugged in values for n2 until the equation equaled itself. For example, I set up the equat...

Ashley Zhu 1A

The energy per photon can be found using the equation E = hv where h is Planck's constant and v is the frequency, which you can find from the given wavelength with the [speed of light = wavelength x frequency] equation.

Ashley Zhu 1A

After finding the energy of a photon using the equation, E = hf (f = frequency), you just need to divide the total energy emitted (11J) by the energy per photon to get the number of photons: 11J/(1.074 x 10^-19 J).

Ashley Zhu 1A

You can also look at the graph they provide in Figure 1A.8, which explains why C) is true. Over the same distance, when the wavelength is longer, the frequency is shorter so there is less change overall than short wavelengths with high frequencies.

Ashley Zhu 1A

Today in discussion we learned that density was related to molarity as they had similar equations (D = m/V and C = n/V) and were given the example: 6.54g (1cm^3/0.7857) = 8.32 cm^3 = 8.32 mL While I understand the calculations and the relationship between cm^3 and mL, I'm not sure where the 6.54 or ...

Ashley Zhu 1A

I have the 7th edition and had the same problem where I spent forever on the problem because it wouldn't balance. I'm pretty sure it's a typo as well because in the solutions manual they also give the answer by using Fe3Br8.

Ashley Zhu 1A

Thank you, that was a great explanation!

Ashley Zhu 1A

Got it, thanks!

Ashley Zhu 1A

I know we haven't talked about this in class yet, but I was doing some basic chemistry review for a life science class and a problem came up for me about orbitals in atoms. A hydrogen atom is simple enough for visualizing--just a spherical shape for the one s-orbital. But for aluminum, would there b...

Ashley Zhu 1A

Relating back to the visual Dr. Lavelle showed us in lecture today of the points on a target, if I only had one point on the bulls eye, I know that would be accurate but would it also be precise? I'm not sure as to whether just one point can be precise or not.

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